Let $₹ 30,000$ be divided into two parts and $₹ x$ and $₹(30,000-x)$
Let it be represented by $1 \times 2$ matrix
$$
[\mathrm{x}(30,000-\mathrm{x})]
$$
Rate of interest is $0.05$ and $0.07$ per rupee.
It is denoted by the matrix $R$ of order $2 \times 1 . R=\left[\begin{array}{l}0 \cdot 05 \\ 0 \cdot 07\end{array}\right]$
(a) $\mathrm{AR}=1800 \quad[\mathrm{x}(30,000-\mathrm{x})]\left[\begin{array}{l}0 \cdot 05 \\ 0 \cdot 07\end{array}\right]$,
$$
[\mathrm{x} \times 0.05+(30,000-\mathrm{x}) \times 0.07]=[1800]
$$
Multiplying by 100
$$
\begin{aligned}
&\text { or } 5 x+21,0000-7 x=180000 \\
&2 x=21,0000-180000 \\
&x=15,000 \\
&\therefore 30,000-x=30,000-15,000=15,000
\end{aligned}
$$
Thus the two parts are 15,000 each.
(b) Then the total interest is $₹ 2000$
$$
\begin{aligned}
&{\left[x(30,000-x)\left[\begin{array}{l}
0 \cdot 05 \\
0 \cdot 07
\end{array}\right]=[2000]\right.} \\
&\Rightarrow[x \times 0.05+(30,000-x) \times 0.07]=[2000]
\end{aligned}
$$
or $\frac{5 x}{100}+\frac{7}{100}(30,000-x)=2000$
Multiplying by 100
$5 x+21,000-7 x=200000$
$2 x=210000-200000$
$$
\begin{aligned}
& x=5000 \\
\therefore \quad & 30,000-x=30,000-5000 \\
&=25,000
\end{aligned}
$$
$\therefore \quad$ Two parts are $₹ 5000$ and 25,000 .